Measuring Fairness We ve seen FOUR methods for tallying votes: Plurality Borda Count Pairwise Comparisons Plurality with Elimination Are these methods reasonable? Are these methods fair? Today we study various fairness criteria. A voting method may or may not satisfy a given fairness criterion. If a voting method violates a fairness criterion, we think of this as a strike against that particular voting method. Paul Koester () MA 111, Voting Theory September 7, 2011 1 / 25
Majority Criterion Majority Criterion If a candidate receives a majority of the first place votes, then this candidate should be deemed the winner of the election. Do all four of our voting methods satisfy the majority criterion? Paul Koester () MA 111, Voting Theory September 7, 2011 2 / 25
Majority Criterion Method of Plurality satisfies the Majority Criterion. Why? Method of Pairwise Comparisons satisfies the Majority Criterion. Why? Paul Koester () MA 111, Voting Theory September 7, 2011 3 / 25
Majority Criterion Method of Plurality satisfies the Majority Criterion. Why? If a candidate receives a majority of the first place votes, this candidate necessarily receives more votes than any other candidate. (i.e., majority implies plurality) Thus, the method of Plurality satisfies the Majority Criterion. Method of Pairwise Comparisons satisfies the Majority Criterion. Why? Paul Koester () MA 111, Voting Theory September 7, 2011 3 / 25
Majority Criterion Method of Plurality satisfies the Majority Criterion. Why? If a candidate receives a majority of the first place votes, this candidate necessarily receives more votes than any other candidate. (i.e., majority implies plurality) Thus, the method of Plurality satisfies the Majority Criterion. Method of Pairwise Comparisons satisfies the Majority Criterion. Why? If a candidate receives a majority of the first place votes, this candidate must have more first place votes than any other candidate. Thus, this candidate will win each of its head to head comparisons (i.e., this candidate is a Condorcet Candidate). Thus, this candidate wins using Pairwise Comparisons. Thus, Pairwise Comparisons satisfies the Majority Criterion. Paul Koester () MA 111, Voting Theory September 7, 2011 3 / 25
Majority Criterion Plurality with Elimination satisfies the Majority Criterion. Why? Paul Koester () MA 111, Voting Theory September 7, 2011 4 / 25
Majority Criterion Plurality with Elimination satisfies the Majority Criterion. Why? During each round, we eliminate a candidate with the fewest first place votes. A Majority Candidate always has more votes than any other candidate, thus the Majority candidate is never eliminated and therefore wins using Plurality with Elimination. Paul Koester () MA 111, Voting Theory September 7, 2011 4 / 25
Majority Criterion AND Borda Count 3 2 A D B B C C D A A receives a majority of the first place votes. Using the 4,3,2,1 point system, who wins? Paul Koester () MA 111, Voting Theory September 7, 2011 5 / 25
Majority Criterion AND Borda Count 3 2 A D B B C C D A A receives a majority of the first place votes. Using the 4,3,2,1 point system, who wins? A: 4 3 + 1 2 = 14 B: 3 3 + 3 2 = 15 C: 2 3 + 2 2 = 10 D: 1 3 + 4 2 = 11 B wins using Borda Count. Paul Koester () MA 111, Voting Theory September 7, 2011 5 / 25
Majority Criterion AND Borda Count Thus, the Borda Count Method fails the Majority Criterion. PLEASE NOTE: This only means that it is possible for the Borda Count Method to fail to deem a Majority Candidate as the winner. In other words, it means that a Majority Candidate could lose in the Borda Count Method. It does NOT mean that a Majority Candidate must lose whenever Borda Count is used. Paul Koester () MA 111, Voting Theory September 7, 2011 6 / 25
Summary Three of our four voting methods pass the Majority Criterion. Borda Count does not satisfy the Majority Criterion. In some sense, then means that Borda Count can produce unfair results. But wait... Paul Koester () MA 111, Voting Theory September 7, 2011 7 / 25
The Condorcet Criterion Recall that a candidate is called a Condorcet Candidate if this candidate wins each head to head competition. Condorcet Criterion If a candidate is a Condorcet Candidate then this candidate should be declared the winner of the election. Pairwise Comparisons satisfies the Condorcet Criterion. Why? Paul Koester () MA 111, Voting Theory September 7, 2011 8 / 25
The Condorcet Criterion Recall that a candidate is called a Condorcet Candidate if this candidate wins each head to head competition. Condorcet Criterion If a candidate is a Condorcet Candidate then this candidate should be declared the winner of the election. Pairwise Comparisons satisfies the Condorcet Criterion. Why? A Condorcet Candidate wins each of its head to head competitions. Therefore, this candidate wins more head to head competitions than any other candidate, and is therefore the winner, using Pairwise Comparisons. Paul Koester () MA 111, Voting Theory September 7, 2011 8 / 25
Condorcet Criterion AND Other Voting Methods Once again... 13 12 6 3 F M B M H B H H B H F F M F M B Last time we saw that B was a Condorcet Candidate. Also Plurality declares M as the winner Borda Count declares H as the winner Plurality with Elimination declares F as the winner Thus, each of these methods fails the Condorcet Criterion. Paul Koester () MA 111, Voting Theory September 7, 2011 9 / 25
Summary Majority Criterion Condorcet Criterion Plurality Satisfies Fails Borda Count Fails Fails Pairwise Comp Satisfies Satisfies Plur. with Elim Satisfies Fails It appears Borda Count is the least fair, and Pairwise Comparisons is the fairest of them all But wait... Paul Koester () MA 111, Voting Theory September 7, 2011 10 / 25
Monotonicity We introduce a third fairness criterion: Monotonicity Suppose a given candidate, say X, wins an election. Then there is an immediate re-election. In this re-election, the only changes in the ballots are changes that favor X. The X should win the re-election as well. Which of our four voting methods satisfy Monotonicity? Paul Koester () MA 111, Voting Theory September 7, 2011 11 / 25
Monotonicity AND Plurality with Elimination Suppose 29 people vote to select one of A, B, and C and the winner will be selected using Plurality with Elimination. The table shows the voters preferences. 7 8 10 4 A B C A B C A C C A B B Plurality with Elimination declares C the winner. Now suppose for some reason the four people that were going to vote (A, C, B) decide at the last minute to change to (C, A, B). C will receive even more first place votes, so certainly this will only solidify C s victory. Right? Paul Koester () MA 111, Voting Theory September 7, 2011 12 / 25
Monotonicity AND Plurality with Elimination This preference schedule shows the modified ballots 7 8 14 A B C B C A C A B Who is eliminated in the first round? Paul Koester () MA 111, Voting Theory September 7, 2011 13 / 25
Monotonicity AND Plurality with Elimination This preference schedule shows the modified ballots 7 8 14 A B C B C A C A B Who is eliminated in the first round? A is eliminated. The new schedule is then 15 14 B C C B Paul Koester () MA 111, Voting Theory September 7, 2011 13 / 25
Monotonicity AND Plurality with Elimination C wins in the original election. In the second election, even more ballots favored C. Yet B ends up winning the re-election. How did this happen? Paul Koester () MA 111, Voting Theory September 7, 2011 14 / 25
Monotonicity AND Plurality with Elimination C wins in the original election. In the second election, even more ballots favored C. Yet B ends up winning the re-election. How did this happen? First, look at the head to head comparisons in the original election: A vs B: A wins A vs C: C wins B vs C: B wins Originally B was eliminated in first round and C triumphed over A in second round. In re-election, C gained first place votes at the cost of A losing first place votes. C will do even better against A. Except, A lost so many first place votes that A is now eliminated in first round. Paradoxically, C was better off in the original election, in which C had fewer first place votes! So Pluarality with Elimination fails the Monotonicity Condition. Paul Koester () MA 111, Voting Theory September 7, 2011 14 / 25
Summary It turns out that Plurality, Borda Count, and Pairwise Comparisons all satisfy the Monotonicity Condition. (The interested student should think about why) Majority Criterion Condorcet Criterion Monotonicity Plurality Satisfies Fails Satisfies Borda Count Fails Fails Satisfies Pairwise Comp Satisfies Satisfies Satisfies Plur. with Elim Satisfies Fails Fails So it appears Pairwise Comparisons is the only fair method for tallying votes... Paul Koester () MA 111, Voting Theory September 7, 2011 15 / 25
Independence of Irrelevant Alternatives We introduce out final fairness criterion: Independence of Irrelevant Alternatives Suppose candidate X is the winner of an election and in a recount one of the non-winning candidates is removed from the ballots, then X should still be the winner of the election. Paul Koester () MA 111, Voting Theory September 7, 2011 16 / 25
IIA An election is between 5 candidates, A, B, C, D, E. Who wins using Borda Count? 2 6 4 1 1 4 4 A B B C C D E D A A B D A C C C D A A E D B D E D B C B E E C E E B A Who wins using Pairwise Comparisons? (First, how many head to head comparisons need to be made?) Paul Koester () MA 111, Voting Theory September 7, 2011 17 / 25
IIA An election is between 5 candidates, A, B, C, D, E. Who wins using Borda Count? 2 6 4 1 1 4 4 A B B C C D E D A A B D A C C C D A A E D B D E D B C B E E C E E B A Who wins using Pairwise Comparisons? (First, how many head to head comparisons need to be made?) Paul Koester () MA 111, Voting Theory September 7, 2011 17 / 25
IIA An election is between 5 candidates, A, B, C, D, E. 2 6 4 1 1 4 4 A B B C C D E D A A B D A C C C D A A E D B D E D B C B E E C E E B A Who wins using Borda Count? A Who wins using Pairwise Comparisons? (First, how many head to head comparisons need to be made?) Paul Koester () MA 111, Voting Theory September 7, 2011 17 / 25
IIA An election is between 5 candidates, A, B, C, D, E. 2 6 4 1 1 4 4 A B B C C D E D A A B D A C C C D A A E D B D E D B C B E E C E E B A Who wins using Borda Count? A Who wins using Pairwise Comparisons? (First, how many head to head comparisons need to be made?) Need to make 4 + 3 + 2 + 1 = 10 head to head comparisons. After a lengthy calculation we see A is the winner. Paul Koester () MA 111, Voting Theory September 7, 2011 17 / 25
IIA Suppose at the last minute, C drops out of the race. 2 6 4 1 1 4 4 A B B C // C// D E D A A B D A C// C // C// D A A E D B D E D B C// B E E C// E E B A The preference schedule simplifies to 2 11 1 4 4 A B D D E D A A A D B D B E B E E E B A Paul Koester () MA 111, Voting Theory September 7, 2011 18 / 25
IIA 2 11 1 4 4 A B D D E D A A A D B D B E B E E E B A Now who wins using Borda Count? Now who wins using Pairwise Comparisons? Paul Koester () MA 111, Voting Theory September 7, 2011 19 / 25
IIA 2 11 1 4 4 A B D D E D A A A D B D B E B E E E B A Now who wins using Borda Count? B Now who wins using Pairwise Comparisons? Paul Koester () MA 111, Voting Theory September 7, 2011 19 / 25
IIA 2 11 1 4 4 A B D D E D A A A D B D B E B E E E B A Now who wins using Borda Count? B Now who wins using Pairwise Comparisons? B Paul Koester () MA 111, Voting Theory September 7, 2011 19 / 25
Recapping, A won the original election (both using Borda Count and Pairwise Comparisons) C, a non-winning candidate drops out of the race. Since C is a non-winning candidate, C s dropping out of the race should be irrelevant. However, C s dropping out of the race causes A to lose to B (both with Borda Count and Pairwise Comparisons) Thus, both Borda Count and Pairwise Comparisons fail the Independence of Irrelevant Alternatives Criteria. Paul Koester () MA 111, Voting Theory September 7, 2011 20 / 25
More on IIA Consider this three person election. 25% 40% 35% A B C C C A B A B Who wins using Plurality with Elimination? Paul Koester () MA 111, Voting Theory September 7, 2011 21 / 25
More on IIA Consider this three person election. 25% 40% 35% A B C C C A B A B Who wins using Plurality with Elimination? A is eliminated in the first round. Then C beats B, 60% to 40%. Paul Koester () MA 111, Voting Theory September 7, 2011 21 / 25
More on IIA Consider this three person election. 25% 40% 35% A B C C C A B A B Who wins using Plurality with Elimination? A is eliminated in the first round. Then C beats B, 60% to 40%. Suppose that B drops out of the race. Then C will inherit B s first place votes, and C will beat A using Plurality with Elimination (which is actually just Plurality since the election is between two candidates) Thus, Plurality with Elimination fails IIA. Paul Koester () MA 111, Voting Theory September 7, 2011 21 / 25
More on IIA Consider this three person election. 25% 40% 35% A B C C A A B C B Who wins using Plurality with Elimination? Paul Koester () MA 111, Voting Theory September 7, 2011 22 / 25
More on IIA Consider this three person election. 25% 40% 35% A B C C A A B C B Who wins using Plurality with Elimination? A is eliminated in the first round. Then C beats B, 60% to 40%. Paul Koester () MA 111, Voting Theory September 7, 2011 22 / 25
Summary Majority Condorcet Monotonicity IIA Plurality Satisfies Fails Satisfies Fails Borda Count Fails Fails Satisfies Fails Pairwise Comp Satisfies Satisfies Satisfies Fails Plur. with Elim Satisfies Fails Fails Fails Thus far, we have not found a fair method of voting! Paul Koester () MA 111, Voting Theory September 7, 2011 23 / 25
So how do people actually count votes? Pairwise comparisons appears to be the best so far (it only fails one of four fairness criteria) Paul Koester () MA 111, Voting Theory September 7, 2011 24 / 25
So how do people actually count votes? Pairwise comparisons appears to be the best so far (it only fails one of four fairness criteria) On the other hand, experts in Social Choice Theory tend to favor Borda Count Method. Why? Paul Koester () MA 111, Voting Theory September 7, 2011 24 / 25
So how do people actually count votes? Pairwise comparisons appears to be the best so far (it only fails one of four fairness criteria) On the other hand, experts in Social Choice Theory tend to favor Borda Count Method. Why? Even though Borda Count fails three of the four fairness criteria, violations are actually quite rare. (Recall: Saying an voting method fails a Fairness Criteria only means the method can be unfair in some elections, not that it is unfair in all elections ) Paul Koester () MA 111, Voting Theory September 7, 2011 24 / 25
The Search for a Fair Voting Method We have seen FOUR voting methods, and we have seen that, in some sense, each may produce unfair results. On the other hand, there are many other methods for tallying votes. Are these other methods unfair? Shouldn t we go searching for a totally fair method? It seemed reasonable to search for a totally fair voting method, until the 1940s when Kenneth Arrow proved Arrow s Impossibility Theorem, which says that EVERY voting method (not just the four we studied in detail) must fail one of the four fairness criteria. Paul Koester () MA 111, Voting Theory September 7, 2011 25 / 25
A few more words on Arrow s Impossibility Theorem Arrow s Impossibily Theorem says that every voting method can produce violations of the fairness criteria. This does NOT mean that every election produces violations of the fairness criteria. All voting methods we discussed produce identical results in elections between two candidates. Furthermore, all two candidate elections satisfy the four fairness criteria. Thus, voting is straightforward and fair in two candidate elections. Voting paradoxes only occur when three or more candidates are involved. Over time, election systems tend toward two party systems in order to eliminate the paradoxes implied by Arrow s Theorem Paul Koester () MA 111, Voting Theory September 7, 2011 26 / 25